Reflect a line using its fixed intersection and one reflected point

The line point is $(10+9\lambda,4-\lambda,2)$. Substitution gives $30+27\lambda-4+\lambda+4=2$, so $\lambda=-1$ and $P=(1,5,2)$.

Choose $A=(10,4,2)$ on l. The normal line is $\mathbf r=(10,4,2)+s(3,-1,2)$. Its foot F satisfies $30+9s-4+s+4+4s=2$, giving $s=-2$, $F=(4,6,-2)$. Thus $A\prime=2F-A=(-2,8,-6)$.

P lies on the mirror plane, so is fixed by reflection. A direction of the image line is $P-A\prime=(3,-3,8)$. Therefore $\mathbf r=(1,5,2)+t(3,-3,8)$, $t\in\mathbb R$. Changing the sign of the direction gives the same line.